Under choice, $c(X)\le d(X)\le w(X)$ and $\chi (X),L(X)\le w(X)$

Choose one point from each nonempty $B\in \mathcal{B}$; the chosen set meets every nonempty open set because $\mathcal{B}$ is a basis, so it is dense and has cardinality at most $\mathrm{\mid }\mathcal{B}\mathrm{\mid }$.

For each $x\in X$, the subfamily $\{B\in \mathcal{B}:x\in B\}$ is a local base at $x$ and has cardinality at most $\mathrm{\mid }\mathcal{B}\mathrm{\mid }$, so every local character, and therefore its supremum $\chi (X)$, is at most $w(X)$.

Given an open cover, choose for each $B\in \mathcal{B}$ that lies in a cover member one such member; these at most $\mathrm{\mid }\mathcal{B}\mathrm{\mid }$ chosen sets still cover $X$, so $L(X)\le w(X)$.

For a pairwise-disjoint family $\mathcal{U}$ of nonempty open sets, choose a point of $D\cap U$ for each $U\in \mathcal{U}$; disjointness makes this assignment injective into $D$, so $\mathrm{\mid }\mathcal{U}\mathrm{\mid }\le d(X)$ and $c(X)\le d(X)$.

Steps 1.1, 1.2, 1.3 and 1.4 give $c(X)\le d(X)\le w(X)$ and $\chi (X),L(X)\le w(X)$.